Compute the last 3 digits of 1^3 + 2^3 + 3^3 + \\ldots + 69^3.
", "hints_md": "There's a formula for this! Try to think of one, or search it up!", "hints_html": "There’s a formula for this! Try to think of one, or search it up!
", "editorial_md": "\nThe formula for the sum of cubes of the first $n$ natural numbers is:\n$$\\displaystyle\\sum_{k=1}^{n} k^3 = (\\frac{n(n+1)}{2})^2$$\n\nGiven that $n=69$, we want to find the sum of cubes from 1 to 69 and then determine the last three digits of that sum. This means we need to compute:$$\\displaystyle\\sum_{k=1}^{69} k^3 = (\\frac{69\\cdot(69+1)}{2})^2$$Which nets us the following:$$\\displaystyle\\sum_{k=1}^{69} k^3 = 2415^2$$Solving further gives a result of $5832225$. From here, because the question is looking for the last 3 digits of this number, the answer is simply $225$.\n\n- botman", "editorial_html": "The formula for the sum of cubes of the first n natural numbers is:
\\displaystyle\\sum_{k=1}^{n} k^3 = (\\frac{n(n+1)}{2})^2
Given that n=69, we want to find the sum of cubes from 1 to 69 and then determine the last three digits of that sum. This means we need to compute:\\displaystyle\\sum_{k=1}^{69} k^3 = (\\frac{69\\cdot(69+1)}{2})^2Which nets us the following:\\displaystyle\\sum_{k=1}^{69} k^3 = 2415^2Solving further gives a result of 5832225. From here, because the question is looking for the last 3 digits of this number, the answer is simply 225.